Nn O.

asked • 10/06/15

the length of a rectangle is 11 yd less than three times the width, and the area of the rectangle is 42 yd^=^2

the length of a rectangle is 11 yd less than three times the width, and the area of the rectangle is 42 yd^=^2 what are the dimensions?

3 Answers By Expert Tutors

By:

Ben K. answered • 10/06/15

Tutor
4.9 (223)

JHU Grad specializing in Math and Science

See tutors like this
See tutors like this
Length = 3*width - 11
 
area = Length * width = 42
 
This is a system of 2 equations and 2 unknowns. How do you solve this, generally speaking?
 
You "solve" for one variable in an equation and plug it into the other.
 
Let's solve for "width" in the second equation.
 
W = 42/L
 
Now we plug this in to the first equation.
 
L = 3 * (42/L) - 11
 
We have 1 equation and 1 unknown, now, which is the length of the rectangle.
 
This is a little messy, but you can isolate (solve) for L with this.
 
Once you have L, go back to the first substitution
 
W = 42/L
 
and you can get W.
 
Good luck!
Upvote • 1 Downvote
More
Report

Michael J. answered • 10/06/15

Tutor
5 (5)

Effective High School STEM Tutor & CUNY Math Peer Leader

See tutors like this
See tutors like this
Let width = x
Let length = 3x - 11
Area = 42 yd2
 
 
Using the formula for area of a rectangle and the variables, we can write an equation.
 
x(3x - 11) = 42
 
3x2 - 11x = 42
 
 
Subtract 42 on both sides of the equation.
 
3x2 - 11x - 42 = 0
 
 
We have a quadratic equation.  Next, we can try to factor using FOIL.
 
(3x + 7)(x - 6) = 0
 
 
Set the factors equal to zero and solve for x.
 
3x + 7 = 0             and             x - 6 = 0
 
x = -7/3                 and             x = 6
 
 
Since we cannot have negative dimensions, we accept the positive value of x:
 
x = 6
 
 
Substitute this value of x into the variables to get the dimensions.
 
width = 6 yards
length = 7 yards
 
 
Upvote • 0 Downvote
More
Report

Sarah Brooke V. answered • 10/06/15

Tutor
4.8 (19)

Professional upbeat tutor experienced in test prep, reading, writing

See tutors like this
See tutors like this
Okay, so you have to come up with an algebraic equation for the length and width. Since they give you length in terms of width, let's call the width "W" and the length "3W-11." If the area of the rectangle is 42, that means the length times the width is 42. So now we have an equation:
 
W(3W-11)=42
 
Distribute and multiply with W to get:
 
3W2-11W=42
 
By logic and process of elimination, we know that W has to be between 5 and 9. Plug some in to see what works. 
 
6 works! (3(6)2-11(6)) ---> (3(36)-66) ---> (108-66)= 42!
 
So now we know that the width is 6. Now we calculate the length by plugging 6 into our length equation:
 
(3(6)-11)=18-11= 7!
 
7x6=42!
 
 
I hope this was helpful. I am available to tutor online most nights and weekends if you need further help with your math (or any other) homework!
 
Sarah Brooke
Upvote • 1 Downvote
More
Report

Ben K.

This is a good method to solve the problem using the "guess and check" method, which is an unusual, but effective way to solve a problem. In "math" speak, it is called the "iterative process". There is nothing wrong at all with this solution.
Report

10/06/15

Michael J.

However, this method is too time consuming when finding the correct solution, especially when there is the possibility of irrational numbers being a solution.   You can guess integers, but not irrational numbers.  The most effective way is to write the equation using the variables that you initially had set up, and then solve it using algebra.  When you performed the guess and check, you defeated the purpose of having the equation.
Report

10/06/15

Ben K.

While everything that you said is true, there is no harm in learning different methods to solve the same problem. In fact, learning different methods is the best way to understand what is actually happening in the problem. Different strokes for different folks...
Report

10/06/15

Sarah Brooke V.

Well, this will probably be the last time I use this part of Wyzant. Thanks, Michael, for your supportive and professional manner. Here I thought we tutors were all in this together. Guess not...
Report

10/06/15

Michael J.

I do not mean any offense.  I just thought that you had a great method using algebra at solving this problem from the start, but seemed to go against it.  The guess and check method you later used is good as Ben stated, but does not always work well in every situation.  I myself have taken criticism for my methods of solving certain problems.  for example
 
When I use vertex form to find the turning point of a function, some will ask me, wouldn't it will be easier to use   h=-b/2a  to get the x coordinate of the vertex, then plug that into the function to get the y-coordinate?
 
Above anything else, we are all in this together and I encourage you to keep participating in this part of Wyzant.  You are needed here more than you know.
Report

10/06/15

Sarah Brooke V.

I just know that some students find FOILing harder than simple logic. I am quite adept at doing math in my head, so it's easy for me to see that it would have to be in between 5 and 7. I also thought that since this is likely a problem for homework, it would be an integer. Some students may find that way quicker and easier than the FOIL method, although that is more reliable and can take less time if a student is competent. However, with that method, one usually comes up with a second answer that doesn't work and may be confused by this. I like to give students a lot of options for solving a problem, which is why I was glad to see you and Ben offering other solutions. I just don't know why you had to critique mine, when all our techniques help you get to the correct answer.
Report

10/06/15

Michael J.

I am not trying to critique your method.  In fact, I never said that your method does not work, rather has some limitations, as all other methods do.  As you saw, I used FOIL to solve the problem because one of the solutions happened to be an integer.  It's just like you said that since this is a homework problem, it would be an integer.  But if the solutions were not integers by any chance, then FOIL would not have worked.   I also agree with you that FOIL is not the easiest to work with because you would have to try and factor huge numbers.  In that case, I would have to use the quadratic formula.
 
x = (-b ± √(b2 - 4ac)) / 2a
 
 
where the quadratic equation is in the form
 
ax2 + bx + c = 0
 
 
As for guess and check, I do find it a good method for integers.  But if you were dealing with the possibilities of irrational numbers, then you would have to consider picking numbers that round off to the nearest integer, and it makes it difficult to use the method.
Report

10/06/15

Sarah Brooke V.

I chose my method knowing it probably an integer as well. It seems that both our methods are equally good in different circumstances to find an integer. If the answer isn't an integer, none of our methods are the easiest, although with a calculator, one could do trial and error, squaring decimals. Also, you probably should have given the quadratic formula in your explanation, since this student may not know it if they can't FOIL. My way is the only foolproof way for someone to find the answer if they haven't learned either. 
Report

10/07/15

Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.
Most questions answered within 4 hours.

OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.